**Schur's Inequality**

For all non-negative real numbers x, y, z and a positive number t,

x^{t}(x-y)(x-z) + y^{t}(y-z)(y-x) + z^{t}(z-x)(z-y)
≥ 0

### Example using Schur's Inequality

Prove: for nonnegative a, b, c, 9(a^{3} + b^{3} + c^{3})
+ 27abc ≥ 2(a + b + c)^{3}

There is a fairly standard approach to symmetric polynomial inequalities.
Write [r,s,t] for the sum of the 6 possible permutations x^{r} y^{s} z^{t} + x^{s} y^{r} z^{t} + ...

Also, the six permutations of (a+b+c)^{3} = a^{3} + b^{3}
+ c^{3} + 3a^{2}b + 3ab^{2} + 3a^{2}c + 3ac^{2}
+ 3b^{2}c + 3bc^{2} + 6abc can be written as

1/2 [3,0,0] + 3[2,1,0] + [1,1,1]

Then our inequality

9(a^{3} + b^{3} + c^{3}) + 27abc ≥ 2(a + b + c)^{3}

reduces (that is, by considering the sum of the six inequalities for the six
permutations of a,b,c) to

9/2 [3,0,0] + 9/2 [1,1,1] ≥ 2( 1/2 [3,0,0] + 3[2,1,0] + [1,1,1])

or

7/2 [3,0,0] + 5/2[1,1,1] ≥ 6[2,1,0]

Now it's clear that [3,0,0] ≥ [1,1,1] (because (a^{3}+b^{3}+c^{3})/3 ≥ abc
by AM≥GM) so it suffices to establish that

[3,0,0] + [1,1,1] ≥ 2 [2,1,0]

But this is precisely

x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y) ≥ 0

which is a case of Schur's inequality, easily established by taking WLOG x ≥ y ≥
z.

### Internet references

### Related pages in this website

Muirhead's Inequality is a
general method for proving symmetric polynomial inequalities

Puzzles

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